1002 Strange fuction

来源：互联网发布：linux下分区调整编辑：程序博客网时间：2024/05/21 17:47

Problem Description

Now, here is a fuction:<br>  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)<br>Can you find the minimum value when x is between 0 and 100.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

Output

Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input

2<br>100<br>200

Sample Output

-74.4291<br>-178.8534

简单题意：

给出一个方程F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x，然后求出指定范围内的最小值。

解题思路形成过程：

这和第一题极其类似，也是用二分法实现解题。但是需要强大的数学知识储备，我也是过了好久，才用到求导这个方法。

感想：

算法，果然还是离不开数学储备。

AC代码：

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
const double n=1e-6;
int t;
double y;
double f1(double x)
{
return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double f2(double x)
{
return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}

using namespace std;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        double ll=0,rr=1e2,mid;
        while(ll+n<=rr)
        {
            mid=(ll+rr)/2;;
            if(f2(mid)>0) rr=mid;
             else ll=mid;
        }
        printf("%.4f",f1(mid));
        cout<<endl;
    }
    return 0;
}

0 0